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Inverse of funtion.

  1. Oct 5, 2012 #1
    I'm reading the first chapter of Topology by Munkres. There we can see:

    "if [itex] f [/itex] is bijective, there exists a function from B to A called the inverse of [itex] f [/itex].

    (...)

    As another situation where care is needed, we note that it is not in general true that

    [itex] f^{-1}(f(A_0) = A_0 [/itex] and [itex] f(f^{-1}(B_0)) = B_0 [/itex]. The relevant rules, which we leave you to check, are the following: If [itex] f: A \rightarrow B [/itex] and [itex] A_0 \subset A [/itex] and [itex] B_0 \subset B [/itex], then

    [itex] A_0 \subset f^{-1}(f(A_0)) [/itex] and [itex] f(f^{-1}(B_0) \subset B_0 [/itex]

    The first inclusion is equality if [itex] f [/itex] is injective and the second inclusion is equality if [itex] f [/itex] is surjective."

    Are there any sense in talking about inverse considering that [itex] f [/itex] is not injective or surjective???
     
  2. jcsd
  3. Nov 2, 2012 #2
    Ah, this is a common misunderstanding, due to a notation issue. If [itex]f : A \rightarrow B[/itex] is any function and [itex]B_0 \subset B[/itex], then by [itex]f^{-1}(B_0)[/itex] people always mean the set [itex]\{ x \in A | f(x) \in B_0 \}[/itex]. Notice that this has meaning even if [itex]f[/itex] is not a bijection. Likewise, if [itex]A_0 \subset A[/itex], we have [itex]f(A) = \{ f(x) | x \in A \}[/itex]. I think that with these definitions you should be able to understand the conclusions made in your book.
     
  4. Nov 2, 2012 #3

    HallsofIvy

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    Oh, dear, oh, dear! Just seeing this question makes me want to hide under the bed!

    The very first time I had to give an explanation of a proof to a class in a topology class, it involved [itex]f^{-1}(A)[/itex] for A a set and I did the whole thing assuming f was invertible!

    If f is a function from set X to set Y, and A is a subset of X, B a subset of Y, then we define f(A) to be the set of all y in Y such that f(x)= y for some x in A and [itex]f^{-1}(B)[/itex] to be the set of all x in X such that f(x) is in B.

    IF f is "one to one and onto", that is, if f is invertibe, then we can show that [itex]f^{-1}(f(A))= A[/itex], but f does not have to be invertible, or even defined on set B for [itex]f^{-1}(B)[/itex] to be defined.

    For example, let f:R=>R be defined by f(x)= x2 and let B= [-4, 4]. Then [itex]f^{-1}(B)= [-2, 2][/itex]. f(2)= f(-2)= 4 so both 2 and -2 are in [itex]f^{-1}(B)[/itex] and for any x between -2 and 2, -4< 0< f(x)< 4, so x is also in [itex]f^{-1}(B)[/itex]. If x< -2 or x> 2, f(x)> 4 so not in [-4, 4].

    Even [itex]f^{-1}([-4, -1])[/itex] is defined. Because there is NO x such that [itex]f(x)= x^2[/itex] is in [-4, -1] so [itex]f^{-1}([-4, -1])[/itex] is the empty set.
     
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